Question

Calculus: Evaluate f(x)=1/x at the specified value of the independent variable and simplify the result.
(f(1+Δx)-f(1))/Δx

Answer 1

\[f(x)=1/x\]
\[\frac{ f(1+Δx)-f(1) }{ Δx }\]

Answer 2

@Loser66

Answer 3

@DebbieG

Answer 4

Um... does anyone know to how to solve this?

Answer 5

So is this a differential?

Answer 6

I just started my calculus class, so we're reviewing with substitutions

Answer 7

Dont you just treat the Δx like we do the "h" in the difference quotient?

(I'm not terribly familiar with differentials... but isn't it just a riff on the difference quotient?)

Answer 8

ok.. well... my best guess is to treat is like a diff quotient. Just use that Δx in place of h.

So you need to use the function:

\(\Large \dfrac{ f(1+Δx)-f(1) }{ Δx }=\dfrac{\dfrac{1}{1+Δx}-1 }{ Δx }\)

and then simplify that....

Answer 9

unless you know what Δx is...if you were given another x value somewhere?

Answer 10

no the problem is as I wrote it

Answer 11

Differentials are weird. I guess it would go something like this?.... (but I'm not terribly confident):

\(\Large \left(\dfrac{1}{1+Δx}-1 \right)\cdot\dfrac{1}{ Δx }\\
\Large \left(\dfrac{1-(1+Δx)}{1+Δx} \right)\cdot\dfrac{1}{ Δx }\\
\Large \left(\dfrac{-Δx}{1+Δx} \right)\cdot\dfrac{1}{ Δx }\\
\Large \dfrac{-1}{1+Δx} \)

Answer 12

but it kind of makes sense, since as Δx->0 that expression goes to -1, and the derivative here \(f\prime(1)=-1\).

Answer 13

I don't understand how you went from step 1 to step 2

Answer 14

@southpaw, I'm sorry, I had to run out unexpectedly....

All I did from step 1 to step 2 was to join everything inside the () over the common den'r....

\(\Large \Large \left(\dfrac{1}{1+Δx}-1 \right)\cdot\dfrac{1}{ Δx }\\
\Large \left(\dfrac{1}{1+Δx}-\dfrac{1+Δx}{1+Δx} \right)\cdot\dfrac{1}{ Δx }\\
\Large \left(\dfrac{1-(1+Δx)}{1+Δx} \right)\cdot\dfrac{1}{ Δx }\\
\Large \left(\dfrac{-Δx}{1+Δx} \right)\cdot\dfrac{1}{ Δx }\\
\Large \dfrac{-1}{1+Δx}\)

Answer 15

alright thank you